A-Level Physics · guided topic map
Oscillations for Cambridge International AS & A Level Physics
Oscillations for A-Level Physics, organized into 1 syllabus topic and 4 mapped concept guides.
- Syllabus topics
- 1
- Mapped concept guides
- 4
- Educational level
- Cambridge International AS & A Level
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17Oscillations
A Level extension
4 guides+
Oscillations
A Level extension
- 01Simple harmonic motionMapped lesson
- 02Energy in simple harmonic motionMapped lesson
- 03Damped oscillationsMapped lesson
- 04Forced oscillations and resonanceMapped lesson
Diagrams
Oscillations as A-Level Physics draws it
The figures from the A-Level Physics practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.
01Fig. 9.1OscillationsA Level
Figure comment
Fig. 9.1A spring hangs vertically from a rigid horizontal support drawn with hatching above it, and a block labelled m is attached to the lower end of the spring. A dashed horizontal line level with the centre of the block is labelled equilibrium position. Two further dashed horizontal lines, one above it and one below it, mark the highest and lowest positions the block reaches, and the distance from the equilibrium line to each of them is marked 4.0 cm.
Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.
Guided questions 5 parts
Reading cue. Each 4.0 cm runs from the equilibrium line to one extreme, so it is the amplitude; the block travels the 8.0 cm between the outer dashed lines twice in each cycle.
aState State the total distance travelled by the block in one complete oscillation between the dashed lines of Fig. 9.1.
Check answer 2 marks
- distance = 4 × amplitude
- = 16 cm (0.16 m)
bCalculate The period of the oscillation is 0.80 s. Calculate the maximum acceleration of the block, and state where on Fig. 9.1 it occurs.
Check answer 3 marks
- ω = 2π/0.80 = 7.85 rad s⁻¹
- a₀ = ω²x₀ = 7.85² × 0.040 = 2.5 m s⁻²
- at the two outer dashed lines, directed towards the equilibrium line
cDetermine The block has a mass of 0.25 kg. Determine the spring constant of the spring and the maximum resultant force on the block.
Check answer 3 marks
- for a mass on a spring ω² = k/m, so k = mω² = 0.25 × 61.7
- k = 15 N m⁻¹
- F = ma₀ = 0.25 × 2.47 = 0.62 N
dDeduce Deduce the extension of the spring when the block is at the equilibrium line marked in Fig. 9.1, and explain why that line is not level with the lower end of the unloaded spring.
Check answer 4 marks
- at the equilibrium position the spring force balances the weight: kx = mg
- x = (0.25 × 9.81) / 15.4 = 0.16 m
- the spring is already stretched by 16 cm in supporting the block, so the equilibrium line lies 16 cm below the unloaded end
- the oscillation takes place about this stretched position, not about the natural length
Transfer challenge
A simple pendulum, for which T = 2π√(L/g), is to swing with the same period, 0.80 s, as the block in Fig. 9.1. Calculate its length, and state what that length has in common with the extension of the loaded spring.
Check answer 3 marks
- L = gT²/4π²
- L = 9.81 × 0.80² / 4π² = 0.16 m
- it is equal to the static extension of the spring, since both are equal to g/ω²